Integration Help: Solving \int \frac{1}{\sqrt{e^{2x}-1}} dx with u-Substitution

[XtremePhysX]

Homework Statement

Find:

Homework Equations

$$\int \frac{1}{\sqrt{e^{2x}-1}} dx$$

The Attempt at a Solution

I tried u=e^x

 

[micromass]

I tried u=e^x

And what did you get?? Please provide a complete attempt.

 

[XtremePhysX]

I think I got it :)

$$\int \frac{dx}{\sqrt{e^{2x}-1}} \\ Let u^2 = e^{2x}-1 \\ \therefore 2u du=2e^{2x}dx \\ udu = e^{2x}dx \\ Now u^2+1=e^{2x} from the substitution so\\ I=\int \frac{udu}{1+u^2} \cdot \frac{1}{u} \\ = \int \frac{du}{1+u^2} \\= \tan^{-1}(\sqrt{e^{2x}-1})+C$$

Micromass, can you give me some high school level or 1st year uni integrals to practice, I did all the ones in my exercise book, I need some challenging integrals, please :)

 

[micromass]

I think I got it :)

$$\int \frac{dx}{\sqrt{e^{2x}-1}} \\ Let u^2 = e^{2x}-1 \\ \therefore 2u du=2e^{2x}dx \\ udu = e^{2x}dx \\ Now u^2+1=e^{2x} from the substitution so\\ I=\int \frac{udu}{1+u^2} \cdot \frac{1}{u} \\ = \int \frac{du}{1+u^2} \\= \tan^{-1}(\sqrt{e^{2x}-1})+C$$

Looks good!

Micromass, can you give me some high school level or 1 year uni integrals to practice, I did all the ones in my exercise book, I need some challenging integrals, please :)

Sure! I'll look for some challenging ones!
If you want to get a good response, you can always make a thread in https://www.physicsforums.com/forumdisplay.php?f=109 asking for some challenging integrals!

Let me look for some goodies.

 

[Dickfore]

Try this one:
$$\int{\frac{dx}{\sqrt{1 + x^4}}}$$

 

[micromass]

Here are some nice ones:

$$\int \sqrt{\tan(x)}dx$$

$$\int e^{\sin(x)}\frac{x\cos^3(x)-\sin(x)}{\cos^2(x)}dx$$

$$\int \frac{1}{1+\sin(x)}dx$$

$$\int \frac{5x^4+1}{(x^5+x^+1)^2}dx$$

 

[XtremePhysX]

Thank you a lot, I will try my best, and then post my solutions here so you guys can check if they are right or wrong.

 

[XtremePhysX]

Here are some nice ones:

(1)$$\int \sqrt{\tan(x)}dx$$

(2)$$\int e^{\sin(x)}\frac{x\cos^3(x)-\sin(x)}{\cos^2(x)}dx$$

(3)$$\int \frac{1}{1+\sin(x)}dx$$

(4)$$\int \frac{5x^4+1}{(x^5+x^+1)^2}dx$$

(1)Answer for the first 1: $$\frac{log(sex(x)(sin(x)+cos(x))(\sqrt{2}\sqrt{tan(x)}+1)))}{2\sqrt{2}}+C$$ The working was very tedious, very hard integral.
(2) Hard :(
(3) Using a Weierstrass substitution, the answer is: $$-\frac{1}{t+1}+C$$
(4) Simple U-Substitution, let u=x^5+x^+1

I couldn't show full working because I'm very tired now :(

 

[XtremePhysX]

Try this one:
$$\int{\frac{dx}{\sqrt{1 + x^4}}}$$

Tried a lot with this one, but it is impossible !
Please show me a short simple way of doing it.

 

[micromass]

Tried a lot with this one, but it is impossible !
Please show me a short simple way of doing it.

I don't think that his integral even has an elementary solution... Maybe he made a typo??

What about this one:

$$\int \frac{1-4x^5}{(x^5-x+1)^2}dx$$

This has a very easy integral and it's obvious once you see it. But it's pretty hard to find.

 

[XtremePhysX]

I don't think that his integral even has an elementary solution... Maybe he made a typo??

What about this one:

$$\int \frac{1-4x^5}{(x^5-x+1)^2}dx$$

This has a very easy integral and it's obvious once you see it. But it's pretty hard to find.

So how do you do this integral? It looks easy but substitutions aren't working :$

 

[Bohrok]

So how do you do this integral? It looks easy but substitutions aren't working :$

What substitutions did you try?

 

[micromass]

Substitutions won't help you. Think about the quotient rules for derivatives.

 

[Millennial]

micromass, I can see the solution because it fits the pattern of the quotient rule but I do not see any way of deriving it using normal integration rules. Partial fractions does not work, substitution is a dead end, integration by parts requires you to integrate an impossible function (or its just me who is not cunning enough to see how to integrate by parts correctly.) Are you aware of any such method?